First, find the derivative of the function. 

.

Then we find that the points where the derivative equals 0 are at  and . Then picking points on the intervals between these critical points and plugging them into the derivative. Where the results are negative, the function is decreasing. Where they are positive, the function in increasing.

Using this method we see that in the on the interval  the function is increasing. On  the function is decreasing. On  the function is increasing. Because the function is increasing to the left of  and decreasing to the right,  is a local maximum. Plugging it into the function, we can find the y value.      

First, find the derivative of the function. 

.

Then we find that the points where the derivative equals 0 are at  and . Then picking points on the intervals between these critical points and plugging them into the derivative. Where the results are negative, the function is decreasing.

Where they are positive, the function in increasing. Using this method we see that in the on the interval  the function is increasing. On  the function is decreasing. On  the function is increasing. Because the function is decreasing to the left of  and increasing to the right,  is a local minimum. Plugging it into the function, we can find the y value.      

 To find the least y-value, we must first find where the minimum of the function is. This is achieved by finding the derivative and then testing values. The derivative of this function is

We then need to factor that and set it equal to 0, which gives us .

We then set each expression equal to 0 to give us our critical points.

This give us critical points at . We then set up a number line and test values on each side of the values. To the left of , you can choose  and plug that into the derivative. We get a positive value. In between the two critical points, you can choose , which gives us a negative value. To the right of , you can choose , which gives a positive value. To find the minimum, you must find the point where the sign changes from negative to positive. That happens at . That is the x-value of the minimum.

To find the y-value, you must plug that x-value into the original function: 

.   

In order to solve this equation, we must first underestand that by taking the derivative of an equation of a graph and setting it equal to zero, we can find the values of  where there are critical points.  Critical points are either local maxima or minima, in order to figure out which you simply plug in numbers before and after that value of  in order to see whether or not the slope increases/decreases as is approaches or leaves that  value.  

In order to take the derivative of equation, the power rule must be applied, .   Taking the derivative of the graph equation, we find that it is .

Setting the equation equal to zero and solving for , we find that the critical points for this graph are present at .  Now we must determine whether these critical points are local minima or maxima.  

In order to determine whether or not they are local minima/maxima, we must determine the slope behavior of the graph around these points.  If the slope is positive towards a critical point and negative away from that critical point, that critical point is a local maxima.  Vice versa for local minima.   is approximately 1.38.   is approximately -1.38.  Therefore we can plug in -2,0,2 into the slope equation in order to determine the behavior of the slope around those points.

Plugging  into the derivative of the graph equation, we find that slope is positive.

Plugging  into the derivative of the graph equation, we find that the slope is positive as well.

Plugging  into the derivative of the graph equation, we find that the slope is positive.

Because the slope is always increasing, this means that there are no local minima in this graph.

To find the local minima of a function, we must find the point(s) at which the first derivative changes from negative to positive.

The first derivative of the function is:

and was found using the following rule:

Now, we must find the critical points, the points at which the first derivative is zero:

Now, we create intervals over which to see whether the first derivative is positive or negative:

On the first interval, the first derivative is negative, while on the second interval it is positive, indicating that the point of the change in sign  is the local minimum.

To find the minimum of a function, find the first derivative. In order to find the derivative of this fuction use the power rule which states, .

Given the function,  and applying the power rule we find the following derivative.

Check the -value at each endpoint and when the first derivative is zero, namely

The smallest value is .

To find the minimum of a function, find the first derivative. In order to find the derivative of this fuction use the quotient rule which states, 

.

Given the function,  and applying the quotient rule we find the following derivative.

 when  and  when , which indicates that has a local minimum at .

To find the relative minimum, you must first find the derivative of the function so that you can find the critical points. When taking the derivative, multiply the exponent by the leading coefficient and then subtract 1 from the exponent: . Then, set that equal to 0 to get your critical points: . Then, test on either side of that point to see what the behavior of the function is. To the left of 2 (plugging in 0, for example), the value is negative. To the right, the value is positive. Since the slope of the function is going from negative to positive at x=2, there is a relative minimum at x=2.

To find the local minima of the function, we must find the x value at which the function's first derivative changes from negative to positive.

First, we must find the first derivative of the function:

The derivative was found using the following rules:

, 

Now, we must find the critical values (where the first derivative is equal to zero) by setting the first derivative equal to zero:

Next, we must create the intervals in which the critical value is the upper and lower limit, respectively:

On the first interval, the first derivative is less than one, and on the second interval, the first derivative is positive (simply plug in any point on the interval into the first derivative function and check the sign). The change from negative to positive occurs at x=3, so  is our answer.   

To find the local minimum, first find the derivative of the function. To take the derivative, multiply the exponent by the coefficient in front of the x term and then decrease the exponent by 1. Therefore, the derivative is: . Then, set that equal to 0 to get your critical point(s): . Put that point on a number line and test values on either side. To the left of 2, the values are negative. To the right of 2, the values are positive. Therefore, the slope goes from negative to positive, indicating there is a local min at x=2.